Calculate Rotational Inertia of Wheel | 15kg Object on Frictionless Incline

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SUMMARY

The discussion focuses on calculating the rotational inertia of a wheel attached to a 15.0-kg object on a frictionless incline. The tension in the rope is determined to be 58.46681 N. The correct formula for calculating the moment of inertia (I) of the wheel is established as I = Fr²/a, where F is the force due to gravity, r is the radius of the wheel (9.0 cm), and a is the acceleration (2.00 m/s²). The user successfully resolves their confusion regarding the torque and inertia calculations.

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A 15.0-kg object is attached to a cord that is wrapped around a wheel of radius r = 9.0 cm (Fig. P8.60). The acceleration of the object down the frictionless incline is measured to be 2.00 m/s2. Assume the axle of the wheel to be frictionless.

p8-60alt.gif


(a) Determine the tension in the rope.
58.46681 N
(b) Determine the moment of inertia of the wheel.
kg·m2
(c) Determine the angular speed of the wheel 2.00 s after it begins rotating, starting from rest.
rad/s
===================================
The tension is correct. I'm stuck on the inertia. I'll use 't' for torque and 'T' for the tension.

t=Iar => I = t/ar = Fr/ar = T/a but this doesn't work. I also tried using the force of gravity and the sigma force in the above equation and none of them are correct.Due tomorrow morning at 8:30 AM EST... :-(
 
Last edited:
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I guess, the formula is t = Ia/r ... ain't it?
 
saket said:
I guess, the formula is t = Ia/r ... ain't it?
... Crap

Yeah that did it. Thanks.

So for anyone searching the formula becomes
I = Fr^2/a
 
Last edited:

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